A rectangle has an area of 32 square inches. It's length is 4 inches less than three times its width. Find the length and width

Question

rectangle area is 32 square inches. its length is 4 inches less than three times its width. Find the length and width. 

Answer 1

If we let the width be x, then the length is 3x - 4.

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|                              |

|                              |  x

|_______________|

       3x - 4

 

x(3x - 4) = 32  <-- Because area is length x width

 

3x^2 - 4x = 32

 

3x^2 - 4x - 32 = 0

 

(3x + 8) ( x -  4) = 0   <--- Factor

 

Solve each, 3x + 8 = 0    OR   x - 4 = 0

                     x = -8/3                  x = 4

 

A negative width does not make sense, so the width must be 4.   Thus, the length 3(4) - 4 = 8.

Answer 2

32=l*w

l=3w-4

Plugging in and factoring, 

(w-4)(3w+8) =0

The only positive value for the width w that solves this equation is w=4, and therefor l=8, and as is easily checked on your fingers l*w=32.

Answer 3

Let width be x, so the length is 3x-4.

Area = length*width = 32 = x(3x-4) 

3x²-4x-32=0

(x-4)(3x+8) =0

x=4 or -8/3(but width shouldn't be negative)

so width is 4 and length is 3x-4=3*4-4=8